In mathematics, the trigonometric moment problem is formulated as follows: given a sequence { c k } k ∈ N 0 {\displaystyle \{c_{k}\}_{k\in \mathbb {N} _{0}}}, does there exist a distribution function σ {\displaystyle \sigma } on the interval [ 0 , 2 π ] {\displaystyle [0,2\pi ]} such that: c k = 1 2 π ∫ 0 2 π e − i k θ d σ ( θ ) , {\displaystyle c_{k}={\frac {1}{2\pi }}\int _{0}^{2\pi }e^{-ik\theta }\,d\sigma (\theta ),} with c − k = c ¯ k {\displaystyle c_{-k}={\overline {c}}_{k}} for k ≥ 1 {\displaystyle k\geq 1}. An affirmative answer to the problem means that { c k } k ∈ N 0 {\displaystyle \{c_{k}\}_{k\in \mathbb {N} _{0}}} are the Fourier-Stieltjes coefficients for some (consequently positive) unique Radon measure μ {\displaystyle \mu } on [ 0 , 2 π ] {\displaystyle [0,2\pi ]} as distribution function.

In case the sequence is finite, i.e., { c k } k = 0 n < ∞ {\displaystyle \{c_{k}\}_{k=0}^{n<\infty }}, it is referred to as the truncated trigonometric moment problem.

Characterization

The trigonometric moment problem is solvable, that is, { c k } k = 0 n {\displaystyle \{c_{k}\}_{k=0}^{n}} is a sequence of Fourier coefficients, if and only if the (n + 1) × (n + 1) Hermitian Toeplitz matrix T = ( c 0 c 1 ⋯ c n c − 1 c 0 ⋯ c n − 1 ⋮ ⋮ ⋱ ⋮ c − n c − n + 1 ⋯ c 0 ) {\displaystyle T=\left({\begin{matrix}c_{0}&c_{1}&\cdots &c_{n}\\c_{-1}&c_{0}&\cdots &c_{n-1}\\\vdots &\vdots &\ddots &\vdots \\c_{-n}&c_{-n+1}&\cdots &c_{0}\\\end{matrix}}\right)} with c − k = c k ¯ {\displaystyle c_{-k}={\overline {c_{k}}}} for k ≥ 1 {\displaystyle k\geq 1}, is positive semi-definite.

The "only if" part of the claims can be verified by a direct calculation. We sketch an argument for the converse. The positive semidefinite matrix T {\displaystyle T} defines a sesquilinear product on C n + 1 {\displaystyle \mathbb {C} ^{n+1}}, resulting in a Hilbert space ( H , ⟨ , ⟩ ) {\displaystyle ({\mathcal {H}},\langle \;,\;\rangle )} of dimensional at most n + 1. The Toeplitz structure of T {\displaystyle T} means that a "truncated" shift is a partial isometry on H {\displaystyle {\mathcal {H}}}. More specifically, let { e 0 , … , e n } {\displaystyle \{e_{0},\dotsc ,e_{n}\}} be the standard basis of C n + 1 {\displaystyle \mathbb {C} ^{n+1}}. Let E {\displaystyle {\mathcal {E}}} and F {\displaystyle {\mathcal {F}}} be subspaces generated by the equivalence classes { [ e 0 ] , … , [ e n − 1 ] } {\displaystyle \{[e_{0}],\dotsc ,[e_{n-1}]\}} respectively { [ e 1 ] , … , [ e n ] } {\displaystyle \{[e_{1}],\dotsc ,[e_{n}]\}}. Define an operator V : E → F {\displaystyle V:{\mathcal {E}}\rightarrow {\mathcal {F}}} by V [ e k ] = [ e k + 1 ] for k = 0 … n − 1. {\displaystyle V[e_{k}]=[e_{k+1}]\quad {\mbox{for}}\quad k=0\ldots n-1.} Since ⟨ V [ e j ] , V [ e k ] ⟩ = ⟨ [ e j + 1 ] , [ e k + 1 ] ⟩ = T j + 1 , k + 1 = T j , k = ⟨ [ e j ] , [ e k ] ⟩ , {\displaystyle \langle V[e_{j}],V[e_{k}]\rangle =\langle [e_{j+1}],[e_{k+1}]\rangle =T_{j+1,k+1}=T_{j,k}=\langle [e_{j}],[e_{k}]\rangle ,} V {\displaystyle V} can be extended to a partial isometry acting on all of H {\displaystyle {\mathcal {H}}}. Take a minimal unitary extension U {\displaystyle U} of V {\displaystyle V}, on a possibly larger space (this always exists). According to the spectral theorem, there exists a Borel measure m {\displaystyle m} on the unit circle T {\displaystyle \mathbb {T} } such that for all integer k ⟨ ( U ∗ ) k [ e n + 1 ] , [ e n + 1 ] ⟩ = ∫ T z k d m . {\displaystyle \langle (U^{*})^{k}[e_{n+1}],[e_{n+1}]\rangle =\int _{\mathbb {T} }z^{k}dm.} For k = 0 , … , n {\displaystyle k=0,\dotsc ,n}, the left hand side is ⟨ ( U ∗ ) k [ e n + 1 ] , [ e n + 1 ] ⟩ = ⟨ ( V ∗ ) k [ e n + 1 ] , [ e n + 1 ] ⟩ = ⟨ [ e n + 1 − k ] , [ e n + 1 ] ⟩ = T n + 1 , n + 1 − k = c − k = c k ¯ . {\displaystyle \langle (U^{*})^{k}[e_{n+1}],[e_{n+1}]\rangle =\langle (V^{*})^{k}[e_{n+1}],[e_{n+1}]\rangle =\langle [e_{n+1-k}],[e_{n+1}]\rangle =T_{n+1,n+1-k}=c_{-k}={\overline {c_{k}}}.} As such, there is a j {\displaystyle j}-atomic measure m {\displaystyle m} on T {\displaystyle \mathbb {T} }, with j ≤ 2 n + 1 < ∞ {\displaystyle j\leq 2n+1<\infty } (i.e. the set is finite), such that c k = ∫ T z − k d m = ∫ T z ¯ k d m , {\displaystyle c_{k}=\int _{\mathbb {T} }z^{-k}dm=\int _{\mathbb {T} }{\bar {z}}^{k}dm,} which is equivalent to c k = 1 2 π ∫ 0 2 π e − i k θ d μ ( θ ) . {\displaystyle c_{k}={\frac {1}{2\pi }}\int _{0}^{2\pi }e^{-ik\theta }d\mu (\theta ).}

for some suitable measure μ {\displaystyle \mu }.

Parametrization of solutions

The above discussion shows that the truncated trigonometric moment problem has infinitely many solutions if the Toeplitz matrix T {\displaystyle T} is invertible. In that case, the solutions to the problem are in bijective correspondence with minimal unitary extensions of the partial isometry V {\displaystyle V}.

See also

Notes

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